A note on badly approximable systems of linear forms over a field of formal series

We prove that the set of badly approximable systems of M linear forms in N variables over the field of formal power series is hyperplane absolute winning. A consequence of this winning property is that the set has full Hausdorff dimension.


Introduction
Metric Diophantine approximation is concerned with the study of approximating irrational numbers by rational numbers from a measure theoretic point of view. Given any irrational number x it is possible to find rational numbers p/q that approximate x to any prescribed accuracy ψ(q) > 0, but as ψ(q) → 0, the rational approximations necessarily have larger and larger denominators. As a consequence of Dirichlet's theorem (1842), for any irrational number x, there exists infinitely many rational numbers p/q such that |x − p/q| < 1/q 2 . Improving this rate of approximation leads to investigate two sets: the set of ψ-approximable numbers and the set of badly approximable numbers. Khintchine (1924) and Jarník (1931) theorems gives a full measure theoretic picture for the size (Lebesgue measure and Hausdorff measure respectively) of ψ-approximable set W (ψ) := {x ∈ R : |x − p/q| < ψ(q) for infinitely many (p, q) ∈ Z × N}.
We refer the reader to [7] for the most recent result in this direction. The numbers x for which there exists a constant c(x) > 0 such that |x − p/q| ≥ c(x)|q| −2 for all p/q ∈ Q are called the badly approximable numbers. From Khintchine's theorem, it follows that the Lebesgue measure of the set of badly approximable numbers is zero. Jarník (1932) proved that the Hausdorff dimension of this set is 1, that is, there are too many badly approximable numbers. Some examples of badly approximable numbers include the quadratic irrationals.
The one dimensional notions extends naturally to higher dimensions, therein, one can consider either simultaneous (approximation by rational points) or dual (approximation by rational hyperplanes) setting. Combining both approaches leads to the theory of approximating systems of linear forms. Historically, the main focus has been on establishing the metrical theory for the set of well and badly approximable systems of linear forms. |q| N /M |qA − p| > 0 denote the set of badly approximable systems of linear forms. Here and throughout, the matrix A is regarded as a point in R MN , the norm |q| = max{|q 1 |, . . . , |q N |} is the supremum norm, and the system of M linear forms in N variables q 1 , . . . , q N is written more concisely as qA.
Wolfgang Schmidt [12] introduced Schmidt's game for showing that the set of badly approximable linear forms is a winning set, thus proving a stronger property than just full Hausdorff dimension. The basic scheme appearing in the definition of a winning set has seen many variants. Some of these include the notions of absolute winning and strong winning sets [11], modified winning sets [6], hyperplane absolute winning, [2], potential winning [4], and Cantor winning [1].
The aim of this paper is to prove that the set of badly approximable linear forms over the field of formal power series is Hyperplane Absolute Winning (HAW). The definition of a Hyperplane game is given below and from it one can easily see that HAW property is in fact stronger than α-winning in a Schmidt sense [2, Proposition 2.3]. The hyperplane game was defined in [2] and it was proved in [3] that systems of badly approximable linear forms over real numbers are hyperplane winning. See also [5, Theorem 1.1] for a proof of Bad(M, N ) to be HAW.
Let F denote the finite field of q = p r elements, where p is a prime and r is a positive integer. We define the field of Laurent series with coefficients from F or the field of formal power series with coefficients from F to be An absolute value · on L can be defined as Under the induced metric d(x, y) = x − y , the space (L, d) is a complete metric space. Furthermore the absolute value satisfies for any x, y ∈ L, x ≥ 0 ∀x, and x = 0 if and only if x = 0, The last property is known as the non-Archimedean property. In fact, equality holds whenever x = y . As we will be working in the finite dimensional vector spaces over L, we need an appropriate extension of the one-dimensional absolute value.
For any x = (x 1 , . . . , x h ) ∈ L h , we define the height of x to be One can easily see that the property (1) is also true for x ∞ .
In L, the polynomial ring F[X] plays a role analogous to the one played by the integers in the field of real numbers. We define the polynomial part of a non-zero element by ⎡ whenever n ≥ 0. When n < 0, the polynomial part is equal to zero. Likewise, the polynomial part of the zero element is itself equal to zero. Using the above definitions, let us define the distance to F[X] h for a point x ∈ L h as Also, for x = (x 1 , . . . , x h ) and y = (y 1 , . . . , y h ) ∈ L h , we define We will make use of the following inequality We identify M × N -matrices with coefficients from L with L MN in the usual way. Matrix products are defined as in the real case. A system of M linear forms in N variables given by A ∈ L MN is called badly approximable if there exists a constant K = K (A) > 0 such that for every q ∈ F[X] N \{0}, Let Bad(M, N ) denotes the set of all badly approximable linear forms A ∈ L MN . In [9], Kristensen proved that the set Bad(M, N ) is of full Hausdorff dimension by proving that this set is winning in terms of the Schmidt (α, β)-game. For further details about the (α, β)game, we refer to Schmidt's paper [13]. For completeness, we state that, Kristensen [8] proved results on Haar measure and Hausdorff dimension of the set of well-approximable matrices over a field of formal series which implies that the Haar measure of Bad(M, N ) is zero. As stated before, HAW is a stronger variant of Schmidt's (α, β)-game, hence a natural question is to investigate the set Bad(M, N ) for HAW property. We prove the following theorem.

Hyperplane Absolute Winning
Let = L MN × R ≥0 . We call the space of formal balls in L MN , where ξ = (C, ρ) ∈ is said to have centre C and radius ρ. Define the map ψ from to the subsets of L MN , assigning a real · ∞ -ball to the abstract one defined above. That is, for ξ = (C, ρ) ∈ , Fix 0 < β < 1 3 and a target set S ⊆ L MN . Hyperplane game is played as follows: • Bob chooses initial ball B 0 ⊆ . • Alice chooses a (MN − 1)-dimensional affine subspace L 1 in L MN and a number 1 , 1 is the 1 -thickening of the subspace L 1 .
• And so on ad infinitum.
We obtain a nested sequence of balls B 0 ⊇ B 1 ⊇ . . . and declare Alice the winner if and only if If Alice has a strategy to win regardless of Bob's play, we say that S is Hyperplane Absolute Winning (HAW).

Proof of Theorem 1.2
To prove this theorem, the plan is to apply strategy developed in [3] to the formal series setting. Let We shall be playing a hyperplane game on L H . For k ∈ N we denote the k th ball chosen by Bob by B(k). Let ρ = ρ(B(0)), and set σ = max{ J ∞ : J ∈ B(0)}. We use boldface lower case letter (x, y, etc.) to denote points in L N and L M . We use boldface upper case letters (P, B, etc.) to denote points in L L . Finally, upper case letters (A, Y, etc.) denote points in L H .
For any matrix where every γ ij is from L, let For Q ∈ L L , let q ∈ L N be the projection of Q onto first N coordinates. Similarly, for P ∈ L L , let p ∈ L M be the projection of P onto the first M coordinates. Set and We notice that a matrix A lies in Bad(M, N ) if and only if there exists a constant c such that for all Q ∈ F[X] L with Q = 0, and It is easy to see that if we fix A ∈ L H and if for each i ∈ N, the system of inequalities (2), Next two lemmas highly rely on the theory of the geometry of numbers. For vector spaces over the field of formal power series the theory was comprehensively developed by Mahler in [10]. A brief summary was given in [9], as well as the equivalent form of these lemmas, we take it from there without a proof. This is also an adaptation of Schmidt's lemmas [13] to the formal series case.
and if for all A ∈ B the system (2), (3) has no solution Q ∈ F[X] L , then the set of all vectors P ∈ F[X] L satisfying (4) with j = i such that there exists A ∈ B satisfying (5) with j = i spans a subspace of L L whose dimension is at most N .
and if for all A ∈ B the system (4), (5) has no solution P ∈ F[X] L , then the set of all vectors Q ∈ F[X] L satisfying (2) with i = j + 1 such that there exists A ∈ B satisfying (3) with i = j + 1 spans a subspace of L L whose dimension is at most M.
Suppose that a game has already been played; for each i ∈ N, let k i ∈ N be the minimal number such that (6) holds with B = B(k i ), and for each j ∈ N, let h j ∈ N be the minimal number such that (7) holds with B = B(h j ). Alice will try to play the game in such a way so that for all A ∈ B(k i ), the system of inequalities (2) and (3) has no solution Q ∈ F[X] L and so that for all A ∈ B(h j ), the system of inequalities (4), (5) has no solution P ∈ F[X] L . The following lemma guarantees that she can always continue to play in such a way.

Lemma 2.3
If R ∈ R is sufficiently large, then the following hold: (i) When the game is at a stage k i , if, for all A ∈ B(k i ), the system of inequalities (2) If u < N , we will add N − u more integer vectors to get a set P = {P 1 , . . . , P N }, which spans a subspace of L L of dimension N . Using P one can easily construct a basis P = {P 1 , . . . , P N } of the same subspace, such that if P j = (P j,1 , . . . , P j,N +M ), P j,i ∈ L, then the matrix has vectors (1, 0 . . . , 0) T , (0, 1, . . . , 0) T , . . . , (0, . . . , 0, 1) T in some order as its N columns (not necessarily the first N columns), while every other element of the matrix has norm P i,j ≤ 1 for i = 1, . . . , N, j = 1, . . . , N + M. This can be done by applying elementary operations (dividing row on the element with the maximum norm will give us '1's and subtracting rows with a suitable coefficient from another ones will give us '0's) on our integer vectors P = {P 1 , . . . , P N }, guaranteed by Lemma 2.1. We will call a set of vectors with these properties a normalized set of vectors. Now suppose that P ∈ S, and let A ∈ B be the corresponding matrix. By the hypothesis, the pair (P, A) do not satisfy the system of inequalities (4), (5) with j = i − 1. But (P, A) do satisfy (5) with j = i, which implies (5) with j = i − 1. Thus they must not satisfy (4) with j = i − 1, i.e. p ∞ ≥ δ T R N (1+(i−1)) .
On the other hand, we may write (5) with j = i as and let D uv = D uv (A, P) be the cofactor of the entry B u · P v in this matrix. We can solve this matrix inequality using Cramer's Rule for every Together with (10) We want now to construct a strategy for Alice such that inequality (12) cannot hold after finite number of steps. This way Alice will avoid all solutions to (4), (5) with j = i, and we can continue with the dual construction. At this point, we introduce some notation. For each v ∈ {0, . . . , N } and for each A ∈ L H , consider the set of v × v minors of the matrix M(A, P) defined by (11). Each minor can be described by such that for any 0 < μ v ≤ ν v and for any normalized set of linearly independent vectors P = {P 1 , . . . , P N } ⊆ L L , Alice can win the following finite game: • Bob plays a closed ball B ⊆ L H satisfying ρ B def = ρ(B) < 1 and max A∈B A ≤ σ .
• Alice and Bob play the hyperplane game until the radius of Bob's ball B v is less than • Alice wins if for all A ∈ B v , we have We will prove this lemma later, but for now let us assume Lemma 2.4 and use it to complete the proof of Lemma 2.3(ii).

Proof of Lemma 2.3 (Part 2) Let ν N > 0 be the number guaranteed by Lemma 2.4 for
Now suppose that the game has progressed to the stage k i , and let ρ k i = ρ(B(k i )). If k i > 0, then since k i is the minimal integer such that (6) is satisfied with B = B(k i ), we have We can ensure that k i > 0 for all i ≥ 0 by requiring that where ρ 0 is the radius of Bob's first ball. In particular, we have This means that this is a valid choice of μ N .
Let P = {P 1 , . . . , P N } be a set of normalized linearly independent vectors, which spans a subspace of L L containing the set S defined by (8). This sets the stage for the finite game described in Lemma 2.4, which is the lemma says Alice can win. Now, the finite game is played until Bob's ball B satisfies in other words, the last ball that Bob plays in the finite game is exactly B(h i ). Since Alice wins the finite game, we have that for every A ∈ B(h i ), (13) holds with v = N . In particular, we have ω = ({1, . . . , N }, {1, . . . , N }) ∈ N and so for all A ∈ B(h i ) D 12 (A) , . . . , D NN (A) ).
That is, (12) is not satisfied for any point A ∈ B(h i ). On the other hand, fixing A ∈ B(h i ), we see by Lemma 2.4 that if there exists a point P ∈ F[X] L satisfying (4) and (5) with j = i, then (12) would also be satisfied, a contradiction. Thus the system of inequalities (4), (5) with j = i has no solution P ∈ F[X] L .
With the proof of Lemma 2.3, we are now in a position to complete the proof of Theorem 1.2.

Completing the proof of Theorem 1.2
Let R ∈ R be chosen large enough so that Lemma 2.3 holds. Note that when i = 0, inequality (2) has no solution Q ∈ F[X] L , and when j = 0, inequality (4) has no solution P ∈ F[X] L . Thus Alice can make dummy moves until stage h 0 , at which point the hypotheses of Lemma 2.3(ii) hold. Then Alice has a strategy to ensure that for all A ∈ B(k 1 ), the system of inequalities (2), (3) with i = 1 has no solution Q ∈ F[X] L . By continuing in this way, Alice ensures that if A is the intersection point of the balls (B(k)) k∈N , then for all i ∈ N, the system of inequalities (2), (3) has no solution Q ∈ F[X] L . This implies that A is badly approximable by what we stated before. Thus the set of badly approximable systems of linear forms over formal series is HAW.

Proof of Lemma 2.4
Finally, the only thing remaining is to prove Lemma 2.4. For this we will need two lemmas, originally developed by Schmidt. To properly formulate them we need some notation. Let ω ∈ v , v ∈ {0, . . . , N }. We define the discrete gradient of D ω (A) to be the vector where E ij ∈ L H is a matrix with value '1' as the ijth entry and zeroes everywhere else. We also define discrete directional derivative of D ω in direction A at point A as Proof Write ω = (I, J ). For any matrix A ∈ L H , we compute Here e 1 , . . . , e N are the standard basis vectors for L L . It follows that that implies equation (14). Now for A ∈ L H , we get the formula and by a similar computation we get (15).
For the next lemma we will also need a notion of a polar (or reciprocal) set of vectors to P. In the formal power series setting this was used by Mahler in [10]. For a set of vectors P = {P 1 , . . . , P N }, P i ∈ L L , linearly independent over L, we call a set T = where δ ij is a Kronecker symbol. A polar set can be explicitly found by  We may choose any value for A e i 0 which lies in L M . In particular, we may let [T j 0 · e M+i ]B i since computation verifies A e i 0 · e M+i = 0 for all i = 1, . . . , N . Now Thus

Whereas,
A ∞ ≤ σ and the lemma follows.
We can now prove Lemma 2.4 by induction on v. When v = 0 the lemma is trivial (by convention we say that max(∅) = 0). Suppose that the lemma has been proven for v − 1, and we want to prove it for v. Let ν v−1 > 0 be given by the induction hypothesis. Fix 0 < μ v−1 ≤ ν v−1 and ν v > 0 to be determined. Suppose that we are given 0 < μ v ≤ ν v and a sequence of normalized vectors P 1 , . . . , P N , and let B be the first ball played by Bob in the finite game. By the induction hypothesis, Alice can play in a way such that if B v−1 is first chosen by Bob satisfying ρ( We must describe how Alice will continue her strategy so as to satisfy (13). We begin with the following observation: We use (14) to bound the following difference: Thus for some ω ∈ v . Let and let F k = F −1 (0). Alice's strategy will be to delete the neighbourhood F (βρ(B v−1 )) k of F k , and then make dummy moves until ρ(B v ) < μ v ρ B . The gradient condition (18) implies that for allÂ ∈ B v−1 \F Next, we have where 3 depends only on σ , M and N . Also by (15) we have Combining (19), (20), (21) and (22) together gives us Applying the trivial bound D ω (Â) ≤ M v (Â) and letting we see that (13)